LUND UNIVERSITY
DEPT. OF STATISTICS
Jakob Bergman
BUSINESS STATISTICS
STAA35
2017-11-15
Exam in Business Statistics (5 ECTS) Saturday 28 October 2017, 08.00–13.00
Permitted aids: Calculator and a collection of formulae distributed with the exam.
To receive full points, every question must be answered with a clearly marked answer and
with a complete and well motivated solution. The total number of points for this exam is 70.
A minimum of 35 points is needed to pass the exam.
Mobile phones (turned off!), briefcases etc. must be placed out of reach during the exam.
MP3-players, smart watches etc. are not allowed. If your calculator can store text (programs),
its memory must be cleared and no text may be stored during the exam.
The cover page must be filled in completely and according to the instructions. It must be submitted whether or not you answer any questions. Write your personal identification number
or birth date on each submitted sheet. Use a new sheet for every question (1, 2, etc.) and use
only one side of the sheet. Write distinctly, illegible answers will not be graded. Do not write
in red. Sort your answers in numerical order.
You are not allowed to leave the exam room until one hour after the exam start.
Questions
1. The random variable X has a binomial distribution with p = 0.8 and n = 10.
(a) What is the expected value of X ?
[2 p]
(b) Calculate P(X = 6).
[3 p]
(c) Calculate P(X ≥ 8).
[5 p]
2. In a survey of students at Gibtsnicht Business School, the students were asked
after their first statistics course exam, whether they went bar hopping the weekend
before the exam or spent the weekend studying, and whether they did well or
poorly on the exam. 60 % of students said they went bar hopping and the others
said they studied for the exam. Of the students who studied for the exam 80 %
did well on the exam, whereas only 30 % of the students who went bar hopping
did well on the exam.
(a) What is the probability that a randomly selected student did well on the
exam?
[4 p]
(b) What is the probability that a randomly selected student went bar hopping
the weekend before the exam if we know that the student did well on the
exam?
[4 p]
(c) Are the events “Did Well on Exam” and “Spent the Weekend Studying”
independent? Why or why not?
[2 p]
1
3. There are 1 000 000 lottery tickets in the “Tack!” scratch card lottery. The following information is available regarding the prizes (in SEK) in the lottery:
Prize
250 000
100 000
10 000
5 000
1 000
500
250
125
50
25
No. of tickets
×
×
×
×
×
×
×
×
×
×
In Total
4
4
8
150
200
1 500
8 880
20 000
12 000
160 000
Total Prize Amount
=
=
=
=
=
=
=
=
=
=
1 000 000
400 000
80 000
750 000
200 000
750 000
2 220 000
2 500 000
600 000
4 000 000
202 746
12 500 000
(a) If you buy a ticket (at random), what is the probability that you have bought
a winning ticket (any prize)?
[2 p]
(b) If you buy a ticket (at random), what is expected amount won on a ticket?
[3 p]
(c) What is the standard deviation of the amount won?∗
[5 p]
4. A certain type of electrical gadget is powered by five LR1234 batteries. The gadget
is operational if at least two of the five batteries are working. It is known that the
life time of a randomly chosen LR1234 battery has normal distribution with a
mean of 720 hours, and it is also known that 15 % of all LR1234 batteries last
more than 746 hours. Five brand new batteries are installed in the gadget. What
is the probability that the gadget will still be operational after 750 hours? [10 p]
5. Studies have shown that the number of cups of coffee that the two co-workers Xerxes and Yvette drink a randomly selected day has the following joint probability
distribution:
Cups of Coffee Consumed by
Probability
0.18
0.65
0.07
0.10
Xerxes
Yvette
0
1
1
2
0
1
2
2
(a) What is the expected number of cups of coffee consumed by Xerxes? [5 p]
(b) Calculate the covariance of the number of cups consumed by Xerxes and the
number of cups consumed Yvette.
[5 p]
∗
If the numbers become too large for your calculator, you divide all the prizes with 1 000 and instead
calculate the expected value and standard deviation in 1 000’s of SEK.
2
6. Small Balls Inc. manufacture red and green balls. The diameter of a ball is normally distributed with a mean of 10 centimetres and a standard deviation of 0.3
centimetres. The balls are sold in boxes, each containing 3 red and 7 green balls.
(a) What is the probability that a randomly selected ball will have diameter less
than 9.8 centimetres?
[2 p]
(b) What is the probability that the mean diameter of the balls in a randomly
selected box is less than 9.8 centimetres?
[4 p]
(c) What is the probability that you well get 2 red and 1 green ball if you randomly select 3 balls from a box?
[4 p]
7. Online customer service is a key element to successful online retailing. According
to a marketing survey, 37.5 % of online customers take advantage of the online
customer service. A random sample of 200 customers is to be selected.
(a) What is probability that the proportion of customers in the sample who take
advantage of online customer service is less than 35 %?
[5 p]
(b) Find a symmetric interval around the population proportion such that there
is a 95 % chance that the proportion of customers in the sample who take
advantage of online customer service will belong to the interval.
[5 p]
Good luck!
Exam Feedback:
Time and venue for the exam feedback is announced in the Live@Lund course web
page and on the notice board at the Department of Statistics.
At the exam feedback you can collect your exam. If you do not attend the exam
feedback you can collect your exam at the department’s exam office after the exam
feedback.
NB The exam feedback is the only opportunity for you to discuss your exam with
the examiner.
3
Solutions to the exam 28 October 2017
1.
(a) E(X ) = np = 10 · 0.8 = 8
(b) P(X = 6) =
10!
6
6!·4! 0.8
· 0.24 = 210 · 0.262144 · 0.0016 = 0.0881
(c) P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
10!
10!
10!
= 8!·2!
0.88 · 0.22 + 9!·1!
0.89 · 0.21 + 10!·0!
0.810 · 0.20
= 0.3020 + 0.2684 + 0.1074 = 0.6778
2. A = Did Well on the Exam
B = Studying for the Exam
(a) and (b) could also be solved using a decision tree
(a) P(A) = P(A and B) + P(A and B0 ) = P(B)P(A|B) + P(B0 )P(A|B0 )
= 0.4 · 0.8 + 0.6 · 0.3 = 0.5
P(A and B0 )
P(B0 )P(A|B0 )
0.6 · 0.3
(b) P(B0 |A) =
=
=
= 0.36
P(A)
P(A)
0.5
(c) No, since P(A|B) = 0.8 6= 0.5 = P(A),
or P(A and B) = 0.32 6= 0.3 = P(A)P(B)
3. A = the ticket is a winning ticket
X = the amount won on a random ticket
(a) P(A) = 202747/1000000 = 0.202747
Total Prize Amount
(b) E(X ) =
= 12500000
1000000 = 12.5 SEK
Total Number of Tickets
(or 0.0125 1 000’s SEK)
or, with aP
little more work
E(X ) = 11
i=1 xi P(X = xi )
= 0 · 0.797254 + 25 · 0.16 + 50 · 0.012 + · · · + 250000 · 0.000004
= 12.5
P
2
(c) V (X ) = 11
i=1 (xi − 12.5) P(X = xi )
= (0 − 12.5)2 · 0.797254 + · · · + (250000 − 12.5)2 · 0.000004
√
√= 295966.25
s = V (X ) = 295966.25 = 544.03 SEK (or 0.54403 1 000’s SEK)
4. X = lifetime of a random battery
X is normal with m = 720
X − 720
746 − 720
>
) = 0.15
P(X > 746) = 0.15 ⇐⇒ P(
s
s
746 − 720
746 − 720
⇐⇒ P(Z >
) = 0.15 ⇐⇒
= 1.0364 [1.04 with Table E.2]
s
s
746 − 720
⇐⇒ s =
= 25.086 [s = 25 if zU = 1.04]
1.0364
Y = number of batteries working after 750 hours
Y is binomial with n = 5 and p = P(X > 750) = 1 − P(X < 750)
750 − 720
= 1 − P(Z <
) = 1 − P(Z < 1.1959) = 1 − 0.8841 = 0.1159
25.086
[0.1151 with s = 25 and Table E.2]
The gadget works if Y ≥ 2
4
P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − [P(Y = 0) + P(Y = 1)]
= 1 − [ 50 0.11590 · 0.88415 + 51 0.11591 · 0.88414 ]
= 1−[0.5402+0.3540] = 1−0.8942 = 0.1058 [0.1045 if p = 0.1151]
5. X = number of cups consumed by Xerxes
Y = number of cups consumed by Yvette
(a) E(X ) = 0 · 0.18 + 1(0.65 + 0.07) + 2 · 0.10 = 0 + 0.72 + 0.2 = 0.92
(b) E(Y ) =P
0 · 0.18 + 1 · 0.65 + 2(0.07 + 0.10) = 0 + 0.65 + 0.34 = 0.99
sXY = 4i=1 [xi − E(Xi )][yi − E(Yi )]P(xi , yi )
= (0 − 0.92)(0 − 0.99)0.18 + (1 − 0.92)(1 − 0.99)0.65
+ (1 − 0.92)(2 − 0.99)0.07 + (2 − 0.92)(2 − 0.99)0.10
= 0.163944 + 0.00052 + 0.005656 + 0.10908 = 0.2792
6.
(a) X = diameter of a random ball
X is normal with m = 10 and s = 0.3
9.8 − 10
P(X < 9.8) = P(Z <
) = P(Z < −0.6667) = 0.2525
0.3
[0.2514 with Table E.2]
(b) X̄ = mean diameter of the balls in a random
√ box
X̄ is normal with m = 10 and sX̄ = 0.3/ 10
9.8 − 10
√ ) = P(Z < −2.1082) = 0.0175
P(X̄ < 9.8) = P(Z <
0.3/ 10
[0.0174 with Table E.2]
(c) Y = number of red balls out of 3 randomly chosen from a box
Y is hypergeometric with n = 3, N = 10, E = 3
3 7
3!
· 7!
3·7
2 1
P(Y = 2) = 10 = 2!·1!10!1!·6! =
= 0.175
120
3!·7!
3
3 72
7 32
126
3 27
or P(RRG) + P(RGR) + P(GRR) = 10
9 8 + 10 9 8 + 10 9 8 = 720 = 0.175
7. p = 0.375 = proportion who use online customer service in the population
n = 200 = sample size
p = proportion who use online customer service in the sample
(a) Since np = 200 · 0.375 = 75 > 5 and n(1 − p) = 200 · 0.625 = 125 > 5
p is approximately
normal with m = p = 0.375 and
p
sp = 0.375 · 0.625/200 = 0.0342
0.35 − 0.375
P(p < 0.35) ≈ P(Z <
) = P(Z < −0.7303) = 0.2326
0.0342
[0.2327 with Table E.2]
(b) Find the value zL and zU corresponding to 2.5 % using Table E.2
pL − 0.375
⇐⇒ pL = 0.375 − 1.96 · 0.0342 = 0.3079
0.0342
pU − 0.375
1.96 =
⇐⇒ pU = 0.375 + 1.96 · 0.0342 = 0.4421
0.0342
−1.96 =
5